SRM 552 - 500 FoxAndFlowerShopDivOne

来源：互联网发布：mab软件编辑：程序博客网时间：2024/05/21 15:45

题目链接：

http://community.topcoder.com/stat?c=problem_statement&pm=11387

题目大意：

一个矩阵中种了两种花，

一个方格里最多种一支花，可以不种。

现在要在矩阵中选取两个不相交的子矩阵，使其中两种花数量之差不超过规定值，且两种花的总数最大。

算法：

这道题是很水的一道枚举题。

两个不相交子矩阵，要不就是没有公共行，要不就是没有公共列。

所以枚举行分界做一遍，再枚举列分界做一遍。

cot1[i][j]代表在列i的左边的子矩阵中，两种花数量之差为i的子矩阵所能具有的最大花总数。

以此类推。

如果以列分界就是分界列左边找一个自矩阵，右边找一个子矩阵，且数量之差要符合要求，枚举即可。

代码：

#include <string>#include <cstdio>#include <iostream>#include <algorithm>#include <sstream>#include <cstdlib>#include <cstring>#include <string>#include <climits>#include <cmath>#include <queue>#include <vector>#include <stack>#include <set>#include <map>#define INF 0x3f3f3f3f#define eps 1e-8using namespace std;const int maxn = 30;int cot1[maxn][maxn * maxn * 2 + 1], cot2[maxn][maxn * maxn * 2 + 1], cot3[maxn][maxn * maxn * 2 + 1], cot4[maxn][maxn * maxn * 2 + 1];int sum1[maxn][maxn], sum2[maxn][maxn];inline int cal1(int x1, int y1, int x2, int y2){    int tmp = sum1[x2][y2];    if (x1)    {        tmp -= sum1[x1 - 1][y2];    }    if (y1)    {        tmp -= sum1[x2][y1 - 1];    }    if (x1 && y1)    {        tmp += sum1[x1 - 1][y1 - 1];    }    return tmp + maxn * maxn;}inline int cal2(int x1, int y1, int x2, int y2){    int tmp = sum2[x2][y2];    if (x1)    {        tmp -= sum2[x1 - 1][y2];    }    if (y1)    {        tmp -= sum2[x2][y1 - 1];    }    if (x1 && y1)    {        tmp += sum2[x1 - 1][y1 - 1];    }    return tmp;}class FoxAndFlowerShopDivOne{public:    int theMaxFlowers(vector <string> flowers, int maxDiff)    {        int ans = -1;        int n = flowers.size();        int m = flowers[0].size();        memset(sum1, 0, sizeof(sum1));        memset(sum2, 0, sizeof(sum2));        memset(cot1, -1, sizeof(cot1));        memset(cot2, -1, sizeof(cot2));        memset(cot3, -1, sizeof(cot3));        memset(cot4, -1, sizeof(cot4));        for (int i = 0; i < n; i ++)        {            for (int j = 0; j < m; j ++)            {                if (flowers[i][j] == 'L')                {                    sum1[i][j] ++;                    sum2[i][j] ++;                }                else if (flowers[i][j] == 'P')                {                    sum1[i][j] --;                    sum2[i][j] ++;                }                if (j)                {                    sum1[i][j] += sum1[i][j - 1];                    sum2[i][j] += sum2[i][j - 1];                }            }            if (i)            {                for (int j = 0; j < m; j ++)                {                    sum1[i][j] += sum1[i - 1][j];                    sum2[i][j] += sum2[i - 1][j];                }            }        }        for (int x1 = 0; x1 < n; x1 ++)        {            for (int y1 = 0; y1 < m; y1 ++)            {                for (int x2 = x1; x2 < n; x2 ++)                {                    for (int y2 = y1; y2 < m; y2 ++)                    {                        cot1[x1][cal1(x1, y1, x2, y2)] = max(cot1[x1][cal1(x1, y1, x2, y2)], cal2(x1, y1, x2, y2));                        cot2[x2][cal1(x1, y1, x2, y2)] = max(cot2[x2][cal1(x1, y1, x2, y2)], cal2(x1, y1, x2, y2));                        cot3[y1][cal1(x1, y1, x2, y2)] = max(cot3[y1][cal1(x1, y1, x2, y2)], cal2(x1, y1, x2, y2));                        cot4[y2][cal1(x1, y1, x2, y2)] = max(cot4[y2][cal1(x1, y1, x2, y2)], cal2(x1, y1, x2, y2));                        if(n==3&&m==3)printf("%d %d %d %d %d %d\n",x1,y1,x2,y2,cal1(x1, y1, x2, y2),cal2(x1, y1, x2, y2));                    }                }            }        }        for (int i = n - 1; i >= 0; i --)        {            for (int k = maxn * maxn - n * m; k <= maxn * maxn + n * m; k ++)            {                if (i < n - 1)                {                    cot1[i][k] = max(cot1[i][k], cot1[i + 1][k]);                }            }        }        for (int i = 0; i < n; i ++)        {            for (int k = maxn * maxn - n * m; k <= maxn * maxn + n * m; k ++)            {                if (i)                {                    cot2[i][k] = max(cot2[i][k], cot2[i - 1][k]);                }            }        }        for (int j = m - 1; j >= 0; j --)        {            for (int k = maxn * maxn - n * m; k <= maxn * maxn + n * m; k ++)            {                if (j < m - 1)                {                    cot3[j][k] = max(cot3[j][k], cot3[j + 1][k]);                }            }        }        for (int j = 0; j < m; j ++)        {            for (int k = maxn * maxn - n * m; k <= maxn * maxn + n * m; k ++)            {                if (j)                {                    cot4[j][k] = max(cot4[j][k], cot4[j - 1][k]);                }            }        }        for (int i = 1; i < n; i ++)        {            for (int k1 = min(maxDiff + 2 * maxn * maxn, 2 * maxn * maxn + n * m); k1 >= 0; k1 --)            {                for (int k2 = maxDiff + 2 * maxn * maxn - k1; k2 >= 0 && k1 + k2 >= 2 * maxn * maxn - maxDiff; k2 --)                {                    if (cot2[i - 1][k2] != -1 && cot1[i][k1] != -1)                    {                        ans = max(ans, cot2[i - 1][k2] + cot1[i][k1]);                    }                }            }        }        for (int j = 1; j < m; j ++)        {            for (int k1 = min(maxDiff + 2 * maxn * maxn, 2 * maxn * maxn + n * m); k1 >= 0; k1 --)            {                for (int k2 = maxDiff + 2 * maxn * maxn - k1; k2 >= 0 && k1 + k2 >= 2 * maxn * maxn - maxDiff; k2 --)                {                    if (cot4[j - 1][k2] != -1 && cot3[j][k1] != -1)                    {                        ans = max(ans, cot4[j - 1][k2] + cot3[j][k1]);                    }                }            }        }        return ans;    }};

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